[BZOJ4756] [Usaco2017 Jan]Promotion Counting（线段树合并）

传送门

 

此题很有意思，有多种解法

1.用天天爱跑步的方法，进入子树的时候ans-query，出去子树的时候ans+query，query可以用树状数组或线段树来搞

2.按dfs序建立主席树

3.线段树的合并

 

前两个都会，于是学习一下线段树的合并。。

道理用文字解释不清。。。直接看代码就能看懂。。

可以脑补出，合并的操作复杂度是logn的，总时间复杂度nlogn

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 100001

int n, m, cnt, size;
int head[N], to[N << 1], next[N << 1], root[N], sum[N * 20], ls[N * 20], rs[N * 20], ans[N], a[N], b[N];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

inline void add(int x, int y)
{
	to[cnt] = y;
	next[cnt] = head[x];
	head[x] = cnt++;
}

inline void insert(int &now, int l, int r, int x)
{
	now = ++size;
	if(l == r)
	{
		sum[now] = 1;
		return;
	}
	int mid = (l + r) >> 1;
	if(x <= mid) insert(ls[now], l, mid, x);
	else insert(rs[now], mid + 1, r, x);
	sum[now] = sum[ls[now]] + sum[rs[now]];
}

inline void merge(int &x, int y)
{
	if(!x || !y)
	{
		x += y;
		return;
	}
	sum[x] += sum[y];
	merge(ls[x], ls[y]);
	merge(rs[x], rs[y]);
}

inline int query(int now, int l, int r, int x, int y)
{
	if(x <= l && r <= y) return sum[now];
	int mid = (l + r) >> 1, ret = 0;
	if(x <= mid) ret += query(ls[now], l, mid, x, y);
	if(mid < y) ret += query(rs[now], mid + 1, r, x, y);
	return ret;
}

inline void dfs(int u)
{
	int i, v;
	for(i = head[u]; i ^ -1; i = next[i])
	{
		v = to[i];
		dfs(v);
		merge(root[u], root[v]);
	}
	ans[u] = query(root[u], 1, m, a[u] + 1, m);
}

int main()
{
	int i, x;
	n = read();
	memset(head, -1, sizeof(head));
	for(i = 1; i <= n; i++) a[i] = b[i] = read();
	for(i = 2; i <= n; i++)
	{
		x = read();
		add(x, i);
	}
	std::sort(b + 1, b + n + 1);
	m = std::unique(b + 1, b + n + 1) - b - 1;
	for(i = 1; i <= n; i++)
		a[i] = std::lower_bound(b + 1, b + m + 1, a[i]) - b;
	for(i = 1; i <= n; i++)
		insert(root[i], 1, m, a[i]);
	dfs(1);
	for(i = 1; i <= n; i++)
		printf("%d\n", ans[i]);
	return 0;
}